QJEEM22Ma67b2: JEE Main 2022 Mathematics - Class 11 JEEM22Ma67b2 Solution: The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where …

Mean $= {{4 + 5 + 6 + 6 + 7 + 8 + x + y} \over 8} = 6$

$\therefore$ $x + y = 12$ ..... (i)

And variance

$= {{{2^2} + {1^2} + {0^2} + {0^2} + {1^2} + {2^2} + {{(x - 6)}^2} + {{(y - 6)}^2}} \over 8}$

$= {9 \over 4}$

$\therefore$ ${(x - 6)^2} + {(y - 6)^2} = 8$ ..... (ii)

From (i) and (ii)

x = 4 and y = 8

$\therefore$ ${x^4} + {y^2} = 320$